Definition 3.1.

A function is said to be a solution of (1.1) if satisfies (1.1).

In what follows one assumes that One needs the following auxiliary result.

Lemma 3.2.

. Let . Then the function defined by

is the unique solution of the boundary value problem

where

Proof.

Let be a solution of the problem (3.2). Then integratingly, we obtain

Hence

where

Now, multiply (3.6) by and integrate over , to get

Thus,

Substituting in (3.6) we have

Therefore

Set Note that

Our first result reads

Theorem 3.3.

Assume that is an -Carathéodory function and the following hypothesis

(A1) There exists such that

holds. If

then the BVP (1.1) has a unique solution.

Proof.

Transform problem (1.1) into a fixed-point problem. Consider the operator defined by

We will show that is a contraction. Indeed, consider Then we have for each

Therefore

showing that, is a contraction and hence it has a unique fixed point which is a solution to (1.1). The proof is completed.

We now present an existence result for problem (1.1).

Theorem 3.4.

Suppose that hypotheses

(H1) The function is an -Carathéodory,

(H2) There exist functions and such that

are satisfied. Then the BVP (1.1) has at least one solution. Moreover the solution set

is compact.

Proof.

Transform the BVP (1.1) into a fixed-point problem. Consider the operator as defined in Theorem 3.3. We will show that satisfies the assumptions of the nonlinear alternative of Leray-Schauder type. The proof will be given in several steps.

Step 1 ( is continuous).

Let be a sequence such that in Then

Since is -Carathéodory and then

Hence

Step 2 ( maps bounded sets into bounded sets in ).

Indeed, it is enough to show that there exists a positive constant such that for each one has .

Let . Then for each , we have

By (H2) we have for each

Then for each we have

Step 3 ( maps bounded set into equicontinuous sets of ).

Let , and be a bounded set of as in Step 2. Let and we have

As the right-hand side of the above inequality tends to zero. Then is equicontinuous. As a consequence of Steps 1 to 3 together with the Arzela-Ascoli theorem we can conclude that is completely continuous.

Step 4 (A priori bounds on solutions).

Let for some . This implies by that for each we have

Then

If we have

Thus

Hence

Set

and consider the operator From the choice of , there is no such that for some As a consequence of the nonlinear alternative of Leray-Schauder type [15], we deduce that has a fixed point in which is a solution of the problem (1.1).

Now, prove that is compact. Let be a sequence in , then

As in Steps 3 and 4 we can easily prove that there exists such that

and the set is equicontinuous in hence by Arzela-Ascoli theorem we can conclude that there exists a subsequence of converging to in Using that fast that is an -Carathédory we can prove that

Thus is compact.